$ E = \left[\begin{array}{rrr}-2 & 0 & 1 \\ 1 & 4 & -2\end{array}\right]$ $ v = \left[\begin{array}{r}3 \\ 5 \\ 4\end{array}\right]$ What is $ E v$ ?
Because $ E$ has dimensions $(2\times3)$ and $ v$ has dimensions $(3\times1)$ , the answer matrix will have dimensions $(2\times1)$ $ E v = \left[\begin{array}{rrr}{-2} & {0} & {1} \\ {1} & {4} & {-2}\end{array}\right] \left[\begin{array}{r}{3} \\ {5} \\ {4}\end{array}\right] = \left[\begin{array}{r}? \\ ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ v$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ v$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ v$ , and so on. Add the products together. $ \left[\begin{array}{r}{-2}\cdot{3}+{0}\cdot{5}+{1}\cdot{4} \\ ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ v$ and add the products together. $ \left[\begin{array}{r}{-2}\cdot{3}+{0}\cdot{5}+{1}\cdot{4} \\ {1}\cdot{3}+{4}\cdot{5}+{-2}\cdot{4}\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{r}{-2}\cdot{3}+{0}\cdot{5}+{1}\cdot{4} \\ {1}\cdot{3}+{4}\cdot{5}+{-2}\cdot{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{r}-2 \\ 15\end{array}\right] $